question_answer

The radii of the two columns in a U - tube are r1 and r2. When a liquid of density p (angle of contact is \[0{}^\circ \]) is filled in it, the level difference of the liquid in the two arms is h. The surface tension of the liquid is:                 (g = acceleration due to gravity)

A)                  \[\frac{\rho gh{{r}_{1}}{{r}_{2}}}{2({{r}_{2}}-{{r}_{1}})}\]                             

B)                  \[\frac{\rho gh({{r}_{1}}-{{r}_{2}})}{{{r}_{2}}{{r}_{1}}}\]

C)                  \[\frac{2({{r}_{1}}-{{r}_{2}})}{\rho gh{{r}_{1}}{{r}_{2}}}\]                             

D)                  \[\frac{\rho gh}{2({{r}_{2}})-{{r}_{1}}}\]

Correct Answer: A

Solution :

                 Rise of liquid in the capillary tube is given by                 \[h=\frac{2T}{r\rho g}\]                                 Suppose, \[{{h}_{1}}\] is the height in tube of radius \[{{r}_{1}}\]and \[{{h}_{2}}\] is the height in tube of radius of \[{{r}_{2}}\]                 So,                          \[{{h}_{1}}=\frac{2T}{{{r}_{1}}\rho g}\]                  ?(i)                 and                        \[{{h}_{2}}=\frac{2T}{{{r}_{2}}\rho g}\]                  ?(ii)                 Level difference of liquid in the two arms is given by                 \[h={{h}_{1}}-{{h}_{2}}=\frac{2T}{{{r}_{1}}\rho g}-\frac{2T}{{{r}_{2}}\rho g}\]                 \[\Rightarrow \]               \[h=\frac{2T}{\rho g}\left[ \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right]\]                 \[\Rightarrow \]               \[h=\frac{2T}{\rho g}\left[ \frac{{{r}_{2}}-{{r}_{1}}}{{{r}_{1}}{{r}_{2}}} \right]\]                 Hence,                  \[T=\frac{h\rho g{{r}_{1}}{{r}_{2}}}{2({{r}_{2}}-{{r}_{1}})}\]